Problem: Let $AB$ be a diameter of a circle centered at $O$.  Let $E$ be a point on the circle, and let the tangent at $B$ intersect the tangent at $E$ and $AE$ at $C$ and $D$, respectively.  If $\angle BAE = 43^\circ$, find $\angle CED$, in degrees.

[asy]
import graph;

unitsize(2 cm);

pair O, A, B, C, D, E;

O = (0,0);
A = (0,1);
B = (0,-1);
E = dir(-6);
D = extension(A,E,B,B + rotate(90)*(B));
C = extension(E,E + rotate(90)*(E),B,B + rotate(90)*(B));

draw(Circle(O,1));
draw(B--A--D--cycle);
draw(B--E--C);

label("$A$", A, N);
label("$B$", B, S);
label("$C$", C, S);
label("$D$", D, SE);
label("$E$", E, dir(0));
dot("$O$", O, W);
[/asy]
Both angles $\angle BAD$ and $\angle CBE$ subtend arc $BE$, so $\angle CBE = \angle BAE = 43^\circ$.  Triangle $BCE$ is isosceles with $BC = CE$, since these are tangent from the same point to the same circle, so $\angle CEB = \angle CBE = 43^\circ$.

Finally, $\angle AEB = 90^\circ$ since $AB$ is a diameter, so $\angle BED = 90^\circ$.  Therefore, $\angle CED = \angle BED - \angle BEC = 90^\circ - 43^\circ = \boxed{47^\circ}$.